# Chemical Bonding

## NCERT Solution

### Part 4

Question 34: Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Answer: Following are the important conditions required for linear combination of atomic orbitals to form molecular ortbials:

• The orbitals present in the valence shell are hybridized.
• The orbitals undergoing hybridization should have almost equal energy.
• Promotion of electron is not a necessary condition prior to hybridization.
• It is not necessary that only half-filled orbitals participate in hybridization. Sometimes, even filled orbitals of valence shell take part in hybridization.

Question 35: Use molecular orbital theory to explain why the Be2 molecule does not exist.

Answer: Electronic configuration of Be is 1s2 2s2.

So, number of electrons in bonding orbitals NB = 4

So, number of electrons in antibonding orbitals NA = 4

Bond order of Be2 =1/2(N_B-N_A)

=1/2(4-4)=0

A negative or zero order means that the molecule is unstable. Hence, molecule Be2 does not exist.

Question 36: Compare the relative stability of the following species and indicate their magnetic properties: O2, O2+, O2- (superoxide), O22- (peroxide)

Answer: Following tables show number of electrons in bonding orbitals and antibonding orbitals in O2

BONDING:

 2s 2p ↑↓ ↑↓ ↑↓ ↑↓

ANTIBONDING

 2s 2p ↑↓ ↑ ↑

NA = 4 and NB = 8

Bond order of O2 =1/2(N_B-N_A)

=1/2(8-4)=2

Following tables show number of electrons in bonding orbitals and antibonding orbitals in O2+

BONDING:

 2s 2p ↑↓ ↑↓ ↑↓ ↑↓

ANTIBONDING

 2s 2p ↑↓ ↑

NA = 3 and NB = 8

Bond order of O2+ =1/2(N_B-N_A)

=1/2(8-3)=2.5

Following tables show number of electrons in bonding orbitals and antibonding orbitals in O2-

BONDING:

 2s 2p ↑↓ ↑↓ ↑↓ ↑↓

ANTIBONDING

 2s 2p ↑↓ ↑ ↑ ↑

NA = 5 and NB = 8

Bond order of O2- =1/2(N_B-N_A)

=1/2(8-5)=1.5

Following tables show number of electrons in bonding orbitals and antibonding orbitals in O22-

BONDING:

 2s 2p ↑↓ ↑↓ ↑↓ ↑↓

ANTIBONDING

 2s 2p ↑↓ ↑↓ ↑ ↑

NA = 6 and NB = 8

Bond order of O22- =1/2(N_B-N_A)

=1/2(8-6)=1

We know that bond dissociation energy is directly proportional to bond order. So, higher bond order is more stable. Now, given molecules or ions can be arranged in decreasing order of stability as follows:

O2+ > O2 > O2- > O22-

Question 37: Write the significance of a plus and a minus sign shown in representing the orbitals.

Answer: Plus sign indicates positive wave function, while minus sign indicates negative wave function of orbitals.

Question 38: Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?

Answer: Ground state configuration of outer shell of P is given below:

 3s 3p 3d ↑ ↓ ↑ ↑ ↑

Excited state configuration of outer shell of P is as follows:

 3s 3p 3d ↑ ↑ ↑ ↑ ↑

Outer shell configuration of PCl5 is as follows:

 3s 3p 3d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓

So, there is sp3d hybridisation in PCl5 molecule. Three P-Cl bonds lie in one plane and make an angle of 120° with each other. These bonds are called equatorial bonds. Out of the two remaining P-Cl bonds, one is above and another is below the equatorial plane and they make an angle of 90° with the equatorial plane. These are called axial bonds. Axial bonds face more repulsion and hence are slightly longer and weaker than equatorial bonds.

Question 39: Define hydrogen bond. Is it weaker or stronger thatn van der Waals forces?

Answer: When a highly electronegative element forms covalent bond with hydrogen atom, the electrons of the covalent bond are shifted towards the more electronegative atom. This results in a partially positively charged hydrogen atom. The partially positively charge hydrogen atom forms a bond with the other more electronegative atom. This bond is called the hydrogen bond and is weaker than the covalent bond. Hydrogen bond is stronger than van der Waals forces because hydrogen bond is considered to be an extreme form of dipole-dipole interaction.

Question 40: What is meant by the term bond order? Calculate the bond order of: N2, O2, O2+ and O2-.

Answer: One half of difference between electrons present bonding and antibonding orbitals is called bond order.

There are 7 electrons in N atom. Following tables show numbers of bonding and antibonding electrons in N2 molecule.

BONDING

 2s 2p ↑↓ ↑↓ ↑↓ ↑↓

ANTIBONDING

 2s 2p ↑↓

NB = 8 and NA = 2

Bond order of N22- =1/2(N_B-N_A)

=1/2(8-2)=3

For O2, O2+ and O2- refer to answer of question 36

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